My intuition tells me that i have to use the rank nullity theorem
Rank(A) + dim(ker(A)) = n
if Rank(A) = n, then that implies that dim(ker(A)) = 0 which means that the ker(A) only has the zero vector… but how can i use this information to prove that A is non-singular/invertible?
This is just a guess on my part, but if Rank(A) is n, then that must mean that the row reduced echelon form(RREF) of A is the identity matrix because the rank of A is the number of NON ZERO rows in the RREF(A), and Rank(A) = n implies that there are no zero rows in RREF(A)
This is as much as i can do on my own
(Please note, i have looked at other posts in Math Exchange that ask the same question, but the answers are too abstract and confusing for me, i just need someone to explain this in steps and ,if possible, to explain this in a more simple way)
Best Answer
If $\ker A=\{0\}$, then the matrix is injective, and since its image is $n$ dimensional, it is surjective, so it is invertible.
Then $$1=\det(I)=\det (AA^{-1})=\det(A)\det(A^{-1})$$
so $\det(A) \neq 0$.